Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 10

1. Let f:A->B and g:B->C be two functions. Show, by direct inference, that 
       (f is injective) and (g is injective) => (g.f is injective)

Sol:
 g.f(x1) = g.f(x2)
=> {Definition of "."}
 g(f(x1)) = g(f(x2))
=> {g is injective}
 f(x1) = f(x2)
=> {f is injective}
 x1 = x2

2. Let f:A->B be a function. Show, by direct inference, that 
       (Exist g:B->A, All x in A, g.f(x) = x) => (f is injective)

Sol:
 f(x1) = f(x2)
=> {g is a function}
 g(f(x1)) = g(f(x2))
=> {Definition of "."}
 g.f(x1) = g.f(x2)
=> {All x in A, g.f(x) = x}
 x1 = x2

3. Let f:A->B be a function. Show, by direct inference, that 
       (Exist g:B->A, All y in B, f.g(y) = y) => (f is surjective)

Sol:
 y in B
=> {All y in B, f.g(y) = y}
 f.g(y) = y
=> {Definition of "."}
 f(g(y)) = y
=> {Let x be the element g(y) in A}
 Exist x in A, f(x) = y

4. Let f:A->B and g:B->C be two functions. Show, by direct inference, that 
       (g.f is bijective) => (f is injective) and (g is surjective)

Sol:
Proving (f is injective)

 f(x1) = f(x2)
=> {g is a function}
 g(f(x1)) = g(f(x2))
=> {Definition of "."}
 g.f(x1) = g.f(x2)
=> {g.f is bijective and so injective}
 x1 = x2

Proving (g is surjective)

 z in C
=> {g.f is bijective and so surjective}
 (Exist x in A, g.f(x) = z)
=> {Definition of "."}
 (Exist x in A, g(f(x)) = z)
=> {Let y be f(x) in B}
 (Exist y in B, g(y) = z)

5. Let f:A->B and g:B->C be two functions. Show, by guessing definitions for 
f and g, that if g and (g.f) are both onto, then f is not necessarily onto.

Sol:
Guess 
 A = {x}, B = {y,y'}, f(x) = y
 C = {z}, g(y) = g(y') = z

In this case, both g and (g.f) are onto, but f is not onto.

6. Let f:A->B be a function that has an inverse f'. Show, by direct inference,
that the function (f'.f) has an inverse and that its inverse is (f'.f) itself.

Sol:
Proving that (f'.f) is 1-1
 f'.f(x1) = f'.f(x2)
=> {definition of "."}
 f'(f(x1)) = f'(f(x2))
=> {f' is 1-1}
 f(x1) = f(x2)
=> {f is 1-1}
 x1 = x2

Proving that (f'.f) is onto
 x in A
=> {f' is inverse of f}
 (f'.f)(x) = x
=> {property of Exist}
 (Exist x1 in A, (f'.f)(x1) = x)

Proving that the inverse of (f'.f) is (f'.f) itself 
 T
=> {Definition of "."}
 ((f'.f).(f'.f))(x) = (f'.f)((f'.f)(x))
=> {(f'.f)(x) = x}
 ((f'.f).(f'.f))(x) = (f'.f)(x)
=> {(f'.f)(x) = x}
 ((f'.f).(f'.f))(x) = x

7. Let f:A->B be a function that has an inverse f', and let B1 be a non-empty 
subset of B. Show, by direct inference, that 
       (f(f'(B1)) sub B1)

Sol:
 y in f(f'(B1))
=> {f(f'(B1)) = B1}
 y in B1

8. Let f:A->B be a function and let A1 and A2 be two non-empty subsets of A. 
Prove, by direct inference, that 
       	  (A1 sub A2) => (f(A1) sub f(A2))

Sol:
 y in f(A1)
=> {Definition of f(A1)}
 Exist x in A1, f(x) = y
=> {A1 sub A2}
 Exist x in A2, f(x) = y
=> {Definition of f(A2)}
 y in f(A2)
 
9. Let R be the set of all real numbers and let f:R->R be the function 
f(x) = x^2 - 3. Show, by guessing, that f is neither injective nor surjective.

Sol:
Proving that f is not injective
Guess: x1 = 2 and x2 = -2. Then, (f(x1) = f(x2)) and (x1 != x2).

Proving that f is not surjective
Note that f(x) >= -3. Thus, there is no x in R such that f(x) = -4.

10. Let R be the set of all real numbers and let f:R->range(f) be the function 
f(x) = 5x + 7. Show that f has an inverse f' and derive the inverse function f'.

Sol:
By choosing the codomain of function f to be range(f), function f is surjective.

Thus, for f to have an inverse f', we need to show that f(x) is injective:

 f(x1) = f(x2)
=> {definition of f}
 (5*x1 + 7) = (5*x2 + 7)
=> {arithmetics}
 (x1) = (x2)

Guess that f'(x) = (x-7)/5. (Hint: f(x) is defined by multiplication followed by 
addition. Thus, f'(x) should be defined by subtraction followed by division.)

Finally, we need to show that f'.f(x) = x. The proof is by direct inference.

 T
=> {Definition of f'.f(x)}
 f'.f(x) = f'(f(x))
         = f'(5x + 7)
         = ((5x + 7) - 7)/5
	 = x

11. Let f:A->B be a function and let A1 and A2 be two subsets of A. Show, by 
guessing values for A, B, f, A1, and A2, that f(A1 ^ A2) is not necessarily 
equal to (f(A1) ^ f(A2)).

Sol:
Guess: A = {1,2,3}
       B = {4,5}
       f(1) = 4, f(2) = 5, f(3) = 4
       A1 = {1,2}
       A2 = {2,3}

f(A1 ^ A2) = {5}
f(A1) ^ f(A2) = {4,5}

Thus, f(A1 ^ A2) != (f(A1) ^ f(A2)).